Question: What is the slope of the line tangent to $f(x) = 2x^{2}+2x-8$ at $x = -3$ ?
Explanation: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(2(x+h)^{2}+2(x+h)-8) - (2x^{2}+2x-8)}{h}$ $ = \lim_{h \to 0} \frac{(2(x^{2}+2x h+h^{2})+2(x+h)-8) - (2x^{2}+2x-8)}{h}$ $ = \lim_{h \to 0} \frac{2x^{2}+4(x h)+2h^{2}+2x+2h-8-2x^{2}-2x+8}{h}$ $ = \lim_{h \to 0} \frac{4(x h)+2h^{2}+2h}{h}$ $ = \lim_{h \to 0} 4x+2h+2$ $ = 4x+2$ $ = (4)(-3)+2$ $ = -10$